Chapter 6 Some Special Discrete Distributions

6.1 THE BERNOULLI DISTRIBUTION 6.2 THE BINOMIAL DISTRIBUTION 6.3 THE GEOMETRIC DISTRIBUTION 6.4 THE POISSON DISTRIBUTION

 6.1.1

6.1 THE BERNOULLI DISTRIBUTION

Bernoulli Trials   Example 1 – The calculation of a payroll check may be correct or incorrect. We define the Bernoulli random variable for this trial so that X = 0 corresponds to a correctly calculated check and X = 1 to an incorrectly calculated one.

Example 2 – A consumer either recalls the sponsor of a T.V. program (X = 1) or does not recall (X = 0)

x p

 Example 3 – In a process for manufacturing spoons each spoon may

X =

1  0

for for a a success

,

failure

.

And, the probability distribution is X P(X = x) 1 p 0 1-p

x

1   0

x

2

P

(

X



x

)   Mathematically, the Bernoulli distribution is given by P(X = x) = p x ( 1- p) 1 - x for x = 1,0    To calculate the mean and variance, 1 Mean,  = E(X) =

x



P x

  0 (

X



x

) Variance,  2 = E(X 2 )  =

x

1   0

x

2

P

(

X



x

)  Note: Since Bernoulli distribution is determined by the value of p, p is the

parameter

of this distribution

6.2 THE BINOMIAL DISTRIBUTION

 6.2.1 The Probability Functions – Consider an experiment which has two possible outcomes, one which may be termed ‘success’ and the other ‘failure’. A binomial situation arises when n independent trials of the experiment are performed , for example – Toss a coin 6 times; – Consider obtaining a head on a single toss as a success and obtaining a tail as a failure; – Throw a die 10 times; – Consider obtaining a 6 on a single throw as a success, and not obtaining a 6 as a failure.

 Example  A coin a biased so that the probability of obtaining a head is . The coin is tossed four times. Find the probability of obtaining exactly two heads.

 Example  An ordinary die is thrown seven times. Find the probability of obtaining exactly three sixes.

 Example  The probability that a marksman hits a target is p and the probability that he misses is q, where q = 1 – p. Write an expression for the probability that, in 10 shots, he hits the target 6 times.

If the probability that an experiment results in a successful outcome is p and the probability that the outcome is a failure is q, where q = 1 – p, and if X is the random variable ‘the number of successful outcomes in n independent trials’, then the probability function of X is given by P(X = x) = for x = 0,1,2,…,n

 Example  If p is the probability of success and q = 1- p is the probability of failure, find the probability of 0,1,2,…,5 successes in 5 independent trials of the experiment. Comment your answer.

In general: The values P(X =x) for x = 0,1,…,n can be obtained by considering the terms in the binomial expansion of (q + p) n , noting that q + p = 1

C

0

q p

0 

C

1

n q n

 1

p

1 

C

2

n q n

 2

p

2  ...



C r n q n



r p r

 ...



C n n q

0

p n

1 = P(X = 0) + P(X =1) + P(X = 2) +…+ P(X= r) +….+P(X = n)

 If X is distribution in this way, we write X  Bin(n,p) where n is the number of independent trials and p is the probability of a successful outcome in one trial – n and p are called the parameters of the distribution.

 Sometimes, we will use b(x; n,p) to represent the probability function when X 

C x n q n



x p x

 Example – The probability that a person supports Party A is 0.6. Find the probability that in a randomly selected sample of 8 voters there are (a) exactly 3 who support Party A, (b) more than 5 who support Party A.

 Example – A box contains a large number of red and yellow tulip bulbs in the ratio 1:3. Bulbs are picked at random from the box. How many bulbs must be picked so that the probability that there is at least one red tulip bulb among them is greater than 0.95?

 6.2.2 Mean and Variance of the Binomial Distribution –   2 = np = npq – Proved it by yourself. :^)

 Example – If the probability that it is find day is 0.4, find the expected number of find days in a week, and the standard deviation.

 Example – The random variable X is such that X  Bin(n,p) and E(X) = 2, Var(X) = . Find the values of n and p, and P(X = 2).



6.2.3 Application

 C.W. Applications of Binomial distributions Throughout this unit, daily lift examples and discussions are the essential features.

– Binomial Distribution Q1The probability that a salesperson will sell a magzine subscription to someone who has been randomly selected from the telephone directory is 0.1. If the salesperson calls 6 individuals this evening, what is the probability that (I) There will be no subscriptions will be sold?

(II) Exactly 3 subscriptions will be sold?

(III) At least 3 subscriptions will be sold?

(IV) At most 3 subscriptions will be sold?

 6.3.1

6.3 THE GEOMETRIC DISTRIBUTION

The Probability Function          A geometric distribution arises when we have a sequence of independent trials, each with a definite probability p of success and probability q of failure, where q = 1 – p. Let X be the random variable ‘the number of trials up to and including the first success’.

Now, P(X = 1) = P(success on the first trial) = p P(X = 2) = P(failure on first trial, success on second) = q p P(X= 3) = ___________________________ P(X = 4) = __________________________ ….

….

P(X = x ) = __________________________

P(X = x) = q x – 1 p, x = 1,2,3,…… where q = 1 – p.

 p is the parameter of the distribution.

 If X is defined in this way, we write X  Geo(p)

 6.3.2 Mean and Variance  = 1

p

and  2 =

q p

2  Proved it by yourself. 

 Example – The probability that a marksman hits the bull’s eye is 0.4 for each shot, and each shot is independent of all others. Find  the probability that he hits the bull’s eye for the first time on his fourth attempt,  the mean number of throws needed to hit the bull’s eye, and the standard deviation,  the most common number of throws until he hits the bull’s eye.

 Example – A coin is biased so that the probability of obtaining a head is 0.6. If X is the random variable ‘the number of tosses up to and including the first head’, find  P(X  4),  P(X > 5),  The probability that more than 8 tosses will be required to obtain a head, given the more than 5 tossed are required.

 Example – In a particular board game a player can get out of jail only by obtaining two heads when she tosses two coins.

 Find the probability that more than 6 attempts are needed to get out of jail.

 What is the smallest value of n if there is to be at least a 90% chance of getting out of jail on or before the n th attempt.

 C.W. Application of Geometric Distribution – 1)The probability that a student will pass a test on any trial is 0.6. What is the probability that he will eventually pass the test on the second trial?

– 2)Suppose the probability that Hong Kong Observatory will make correct daily whether forecasts is 0.8. In the coming days, what is the probability that it will make the first correct forecast on the fourth day?

6.4 THE POISSON DISTRIBUTION

 [see textbook]   Example Verify that if X  Po(  ), then X is a random variable.

  Example If X  Po(  ) find (a) E(X), (b) E(X2), (c) Var(X).

 From above example , we can conclude that the MEAN and VARIANCE of the Poisson distribution are  and  respectively.

C.W. Application of Poisson Distribution

– 1)The average number of claims per day made to the Insurance Company for damage or losses is 3.1. What is the probability that in any given day     fewer than 2 claims will be made?

exactly 2 claims will be made?

2 or more claims will be made?

more than 2 claims will be made?

– 2) Based on past experience, 1% of the telephone bills mailed to house-holds in Hong Kong are incorrect. If a sample of 10 bills is selected, find the probability that at least one bill will be incorrect. Do this using two probability distributions (the binomial and the Poisson) and briefly compare and explain your result.